Integrand size = 29, antiderivative size = 96 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^2 b^3 d}-\frac {a \sin (c+d x)}{b^2 d}+\frac {\sin ^2(c+d x)}{2 b d} \]
-csc(d*x+c)/a/d-b*ln(sin(d*x+c))/a^2/d+(a^2-b^2)^2*ln(a+b*sin(d*x+c))/a^2/ b^3/d-a*sin(d*x+c)/b^2/d+1/2*sin(d*x+c)^2/b/d
Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {2 \csc (c+d x)}{a}-\frac {2 b \log (\sin (c+d x))}{a^2}+\frac {2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^2 b^3}-\frac {2 a \sin (c+d x)}{b^2}+\frac {\sin ^2(c+d x)}{b}}{2 d} \]
((-2*Csc[c + d*x])/a - (2*b*Log[Sin[c + d*x]])/a^2 + (2*(a^2 - b^2)^2*Log[ a + b*Sin[c + d*x]])/(a^2*b^3) - (2*a*Sin[c + d*x])/b^2 + Sin[c + d*x]^2/b )/(2*d)
Time = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3316, 27, 522, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^2 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {\csc ^2(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\csc ^2(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b^2 (a+b \sin (c+d x))}d(b \sin (c+d x))}{b^3 d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (-\frac {\csc (c+d x) b^3}{a^2}+\frac {\csc ^2(c+d x) b^2}{a}+\sin (c+d x) b-a+\frac {\left (a^2-b^2\right )^2}{a^2 (a+b \sin (c+d x))}\right )d(b \sin (c+d x))}{b^3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b^4 \log (b \sin (c+d x))}{a^2}+\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^2}-\frac {b^3 \csc (c+d x)}{a}-a b \sin (c+d x)+\frac {1}{2} b^2 \sin ^2(c+d x)}{b^3 d}\) |
(-((b^3*Csc[c + d*x])/a) - (b^4*Log[b*Sin[c + d*x]])/a^2 + ((a^2 - b^2)^2* Log[a + b*Sin[c + d*x]])/a^2 - a*b*Sin[c + d*x] + (b^2*Sin[c + d*x]^2)/2)/ (b^3*d)
3.14.14.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.59 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {-\frac {-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \sin \left (d x +c \right )}{b^{2}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} a^{2}}}{d}\) | \(90\) |
| default | \(\frac {-\frac {-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \sin \left (d x +c \right )}{b^{2}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} a^{2}}}{d}\) | \(90\) |
| parallelrisch | \(\frac {4 \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+\left (-4 a^{4}+8 a^{2} b^{2}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{4}-2 a \,b^{3} \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-4 b \left (a \sin \left (d x +c \right )+\frac {b \cos \left (2 d x +2 c \right )}{4}-\frac {b}{4}\right ) a^{2}}{4 a^{2} b^{3} d}\) | \(150\) |
| risch | \(-\frac {i a^{2} x}{b^{3}}+\frac {2 i x}{b}-\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 b d}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b d}-\frac {2 i a^{2} c}{b^{3} d}+\frac {4 i c}{b d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{3} d}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b d}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{2} d}\) | \(276\) |
| norman | \(\frac {\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d}+\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d}-\frac {1}{2 a d}-\frac {\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}-\frac {2 \left (a^{2}+b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{2} d}-\frac {2 \left (a^{2}+b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{2} d}-\frac {\left (4 a^{2}+3 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{2} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{2} b^{3} d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}-\frac {\left (a^{2}-2 b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}\) | \(289\) |
1/d*(-1/b^2*(-1/2*sin(d*x+c)^2*b+a*sin(d*x+c))-1/a/sin(d*x+c)-1/a^2*b*ln(s in(d*x+c))+1/b^3*(a^4-2*a^2*b^2+b^4)/a^2*ln(a+b*sin(d*x+c)))
Time = 0.43 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.39 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {4 \, a^{3} b \cos \left (d x + c\right )^{2} - 4 \, b^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 4 \, a^{3} b - 4 \, a b^{3} + 4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) - {\left (2 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} - a^{2} b^{2}\right )} \sin \left (d x + c\right )}{4 \, a^{2} b^{3} d \sin \left (d x + c\right )} \]
1/4*(4*a^3*b*cos(d*x + c)^2 - 4*b^4*log(1/2*sin(d*x + c))*sin(d*x + c) - 4 *a^3*b - 4*a*b^3 + 4*(a^4 - 2*a^2*b^2 + b^4)*log(b*sin(d*x + c) + a)*sin(d *x + c) - (2*a^2*b^2*cos(d*x + c)^2 - a^2*b^2)*sin(d*x + c))/(a^2*b^3*d*si n(d*x + c))
Timed out. \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}} - \frac {b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{b^{2}} + \frac {2}{a \sin \left (d x + c\right )} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b^{3}}}{2 \, d} \]
-1/2*(2*b*log(sin(d*x + c))/a^2 - (b*sin(d*x + c)^2 - 2*a*sin(d*x + c))/b^ 2 + 2/(a*sin(d*x + c)) - 2*(a^4 - 2*a^2*b^2 + b^4)*log(b*sin(d*x + c) + a) /(a^2*b^3))/d
Time = 0.38 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac {b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{b^{2}} - \frac {2 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{a^{2} \sin \left (d x + c\right )} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{3}}}{2 \, d} \]
-1/2*(2*b*log(abs(sin(d*x + c)))/a^2 - (b*sin(d*x + c)^2 - 2*a*sin(d*x + c ))/b^2 - 2*(b*sin(d*x + c) - a)/(a^2*sin(d*x + c)) - 2*(a^4 - 2*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/(a^2*b^3))/d
Time = 11.85 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.43 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,{\left (a^2-b^2\right )}^2}{a^2\,b^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^2-2\,b^2\right )}{b^3\,d}-\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+b^2\right )}{b^2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^2+b^2\right )}{b^2}-\frac {4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{b}+1}{d\,\left (2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]
(log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(a^2 - b^2)^2)/( a^2*b^3*d) - tan(c/2 + (d*x)/2)/(2*a*d) - (log(tan(c/2 + (d*x)/2)^2 + 1)*( a^2 - 2*b^2))/(b^3*d) - (b*log(tan(c/2 + (d*x)/2)))/(a^2*d) - ((2*tan(c/2 + (d*x)/2)^2*(2*a^2 + b^2))/b^2 + (tan(c/2 + (d*x)/2)^4*(4*a^2 + b^2))/b^2 - (4*a*tan(c/2 + (d*x)/2)^3)/b + 1)/(d*(2*a*tan(c/2 + (d*x)/2) + 4*a*tan( c/2 + (d*x)/2)^3 + 2*a*tan(c/2 + (d*x)/2)^5))